Problem: Simplify the following expression: $y = \dfrac{-4x^2- 3x+10}{x + 2}$
First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-4)}{(10)} &=& -40 \\ {a} + {b} &=& &=& {-3} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-40$ and add them together. Remember, since $-40$ is negative, one of the factors must be negative. The factors that add up to ${-3}$ will be your ${a}$ and ${b}$ When ${a}$ is ${5}$ and ${b}$ is ${-8}$ $ \begin{eqnarray} {ab} &=& ({5})({-8}) &=& -40 \\ {a} + {b} &=& {5} + {-8} &=& -3 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-4}x^2 +{5}x) + ({-8}x +{10}) $ Factor out the common factors: $ x(-4x + 5) + 2(-4x + 5)$ Now factor out $(-4x + 5)$ $ (-4x + 5)(x + 2)$ The original expression can therefore be written: $ \dfrac{(-4x + 5)(x + 2)}{x + 2}$ We are dividing by $x + 2$ , so $x + 2 \neq 0$ Therefore, $x \neq -2$ This leaves us with $-4x + 5; x \neq -2$.